If 1-sin 2-3 sin-cos- then prove that tan-1 or 1-2-

1+ sin ^2 θ = 3 sin θ cos θ Divide both sides of the equation with cos^2θ 1/cos^2θ + sin^2θ/cos^2θ = 3 sinθ/cosθ sec^2θ + tan^2θ = 3tanθ 1 + tan^2θ + ta ...

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Standard X

Mathematics

Trigonometric Ratios

If 1+sin 2θ=3...

Question

If $1 + s i n^{2} θ = 3 s i n θ c o s θ$ then prove that $t a n θ = 1 o r \frac{1}{2}$ .

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Solution

$1 + s i n^{2} θ = 3 s i n θ c o s θ Divide both sides of the equation with c o s^{2} θ \frac{1}{c o s^{2} θ} + \frac{s i n^{2} θ}{c o s^{2} θ} = \frac{3 s i n θ}{c o s θ} s e c^{2} θ + t a n^{2} θ = 3 t a n θ 1 + t a n^{2} θ + t a n^{2} θ = 3 t a n θ 1 + 2 t a n^{2} θ - 3 t a n θ = 0 Substitute t a n θ = a 2 a^{2} - 3 a + 1 = 0 Solve the quadratic equation to find out the roots. 2 a^{2} - 2 a - a + 1 = 0 2 a (a - 1) - 1 (a - 1) = 0 (2 a - 1) (a - 1) = 0 2 a - 1 = 0 and (a - 1) = 0 2 a = 1 and a = 1 a = \frac{1}{2} and a = 1 Hence a = t a n θ t a n θ = \frac{1}{2} and t a n θ = 1$

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Similar questions

Q. Question 4
If

1 + s i n^{2} θ = 3 s i n θ . c o s θ

, then prove that

t a n θ = 1

\frac{1}{2}

State true or false.

(1 + {sin}^{2} θ) = 3 sin θ cos θ

then

tan θ = - 1 o r \frac{1}{2} .

Q. Prove that

{(tan θ + \frac{1}{cos θ})}^{2} + {(tan θ - \frac{1}{cos θ})}^{2} = 2 (\frac{1 + {sin}^{2} θ}{1 - {sin}^{2} θ})

Q. Prove that :

\frac{s i n (90^{\circ} - θ) s i n θ}{t a n θ} - 1 = - s i n^{2} θ

Q. If

1 + {sin}^{2} θ = 3 sin θ cos θ

then the solution set in

[0, \frac{π}{2}]

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If 1+sin2θ=3sinθcosθ then prove that tanθ=1or12.

If $1 + s i n^{2} θ = 3 s i n θ c o s θ$ then prove that $t a n θ = 1 o r \frac{1}{2}$ .